Q.A random sample of 225 customer bills has a mean value of Rs. 200 with a population standard deviation of Rs. 30. Construct a 99% confidence interval for the true mean bill value.

A 99% confidence level demands more certainty than 95%, so it uses a larger Z-value and produces a wider interval around the same point estimate. …

Given: $\bar{x} = 200$, $\sigma = 30$, $n = 225$, confidence level = 99% $\Rightarrow Z_{\alpha/2} = 2.576$.

Step 1 - standard error:

Step 2 - margin of error:

Step 3 - confidence interval:

$200 \pm 5.152 = (194.848,\ 205.152)$ …

Compare with a 95% interval for the same data: SE=2, margin = 1.96x2 = 3.92, giving (196.08, 203.92) - visibly narrower than the 99% interval, confirming th …

Using 1.96 (the 95% value) out of habit instead of the correct 99% value 2.576 is the most common slip - always re-check the Z …